a series RLC circuit has a peak current of 1 A with a frequency of 54 kHz. if the resistance of the circuit is 51 kHz, the capacitance of the circuit is 19 uF and the inductance of the circuit is 25 uF, determine the average power of the circuit.

Answer 1

assume you mean r = 51 k Ohm

i = 1 sin 108 pi t

Vr = i r = 51*10^3 sin 108 pi t

Vl = L di/dt = 25*10^-6 * 108pi cos 108 pi t

dVc/dt = (1/C) i
so
Vc = (19*10^-6)^-1 [-(1/108pi) cos 108 pi t

V = Vr + Vl + Vc

Vl + Vc = [.00848 - 155]cos 108 pi t
(I suspect you have some units wrong)
V = 51*10^3 sin 108 pi t - 155 cos 108 pi t

I am sure your input data is incorrect but now multiply i times V and integrate over a cycle and divide by the period (multiply by the frequency)
In fact only the sin*sin term will matter in the end because sin * cos averages to 0 and the average of sin^2 is 1/2

Answer 2

I = 0.707 * 1 = 0.707A rms.
P = I^2*R = (0.707)^2 * 51 = 25.5 kW.